Statement

The inverse map in a group, i.e. the map sending any element of the group, to its inverse element, is an involutive map, in the sense that it has the following two properties:

It satisfies the reversal law:

Applying it twice sends an element to itself:

It fact, both these are true in the greater generality of a monoid, under the condition that all the s have two-sided inverse (note: we still need a monoid to guarantee that two-sided inverses, when they exist, are unique).

Proof

Proof of reversal law

In order to show that the element is a two-sided inverse of , it suffices to show that their product both ways is the identity element. Consider first the product:

Due to associativity, we can drop the parentheses and we get:

Now, consider the middle product . This is the identity element, and since the identity element has no effect on the remaining product, it can be removed, giving the product:

We now repeat the argument with the middle product and cancel them. Proceeding this way, we are able to cancel all terms and eventually get the identity element.

A similar argument follows for the product the other way around:

Thus, the elements are two-sided inverses of each other.

Note: In fact, it suffices to check only one of the two inverse conditions, i.e., check only that the first product is the identity element. This is because, in a group, every element has a two-sided inverse. Further, equality of left and right inverses in monoid forces any one-sided (left or right) inverse to be equal to the two-sided inverse.

Proof for applying it twice

This is direct from the definition. let . Then, by the inherent symmetry in the definition of inverse element, we see that .

More explicitly, if , that means that . But this is precisely the condition for stating that .